$A linear ODE$

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$A linear ODE$

Find the $\textbf{general}$ solution of

$y'= \left (\begin{matrix} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right ) \cdot y$

and a solution with the initial value $y(0) = \left ( \begin{matrix} 1 \\ -1 \\ 1 \end{matrix} \right )$.

Ordinary Differential Equations Differential Equations

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To find the eigenvectors, we solve for the roots of the characteristic polynomial:
$$(1-\lambda)(\lambda^2-1) - (\lambda - 1) + (\lambda - 1) = 0 \implies \lambda_1 = \lambda_2 = 1, \lambda_3 = -1$$.

Solving for the eigenvector associated to $\lambda_1 = \lambda_2 = 1$ yields $v_1 = (2, -1, 1)$. Solving for the eigenvector associated to $\lambda_3 = -1$ yields $v_3 = (0, -1, 1)$. Now to find the missing generalized eigenvector one must solve $(A - I)v_2 = v_1$. A solution of this equation is given by $v_2 = (-1, 2, 0)$. Thus the general solution is
$$c_1 v_1 e^t + c_2(v_1 te^t + v_2 e^t) + c_3v_3e^{-t}$$
where $c_i$ are constants.

For the solution where $y(0) = (1, -1, 1)$, one notes that $y(0) = (v_1 + v_3)/2$. Hence $c_1 = 1/2$, $c_2 = 0$ and $c_3 = 1/2$.

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$A linear ODE$

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